用户: Yao/泛函分析续论

Banach algebra	Banach space with $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$
involution	$a^{\ast \ast }=a,(ab{)}^{\ast }=b^{\ast }{a}^{\ast },(\alpha a+b{)}^{\ast }=\bar{\alpha }{a}^{\ast }+b^{\ast }$
$C^{\ast }$ algebra	Banach algebra with $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$
hermitian elements $\mathrm{Re}\mathcal{A}$	$a^{\ast }=a$
maximal ideal space of abelian $\mathcal{A}$	$\Delta =\{$all nonzero homomorphisms $h:\mathcal{A}\to \mathbb{C}\}$
Gelfand transform of $a$	$\hat{a}:\Delta \to \mathbb{C},h\mapsto h(a)$
$C^{\ast }(a)$	the $C^{\ast }$ subalgebra generated by $a$ and $1$
positive elements in $\mathcal{A}$	$a\in \mathrm{Re}\mathcal{A}$ st $\sigma (a)\subseteq {\mathbb{R}}_{+}$
state space $\Sigma (\mathcal{A})$	all positive functionals of norm $1$

Prop. The correspondence $h\mapsto \ker h$ between homomorphisms and maximal ideals is bijective.

证明. For a maximal ideal $M$, let $\pi :\mathcal{A}\to \mathcal{A}/M$, and suppose that $\pi (a)$ is not invertible in $\mathcal{A}/M$. Note that $M\subseteq$ the ideal $I={\pi }^{-1}(\pi (⟨a⟩))={\pi }^{-1}(⟨\pi (a)⟩)$, then $M=I$, hence $\pi (⟨a⟩)=\pi (I)=0$, so $\pi (a)=0$. Since $\sigma (\pi (a))$ is nonempty, $\pi (a)$ is a number. Therefore $h:\mathcal{A}\to \mathcal{A}/M\simeq \mathbb{C}$ is such a homo that $\ker h=M$.

For a homo $h$, $A/\ker h\simeq \mathbb{C}$, so $\ker h$ is maximal.

$\square$

Prop. $\sigma (a)=\{h(a):h\in \Delta \}$.

证明. $a-h(a)\in \ker h$, so $h(a)\in \sigma (a)$; for $\lambda \in \sigma (a)$, a maximal ideal containing $a-\lambda$ is of the form $\ker h$, so $h(a)=\lambda$.

$\square$

Prop. The kernel of $\Lambda :\mathcal{A}\to C(\Delta ),a\mapsto \hat{a}$ is the intersection of all maximal ideals of $\mathcal{A}$, and $\Vert \hat{a}{\Vert }_{\infty }=r(a)$.

证明. The former obvious, the latter by the former proposition.

$\square$

Prop. Let $\mathcal{A}$ be a $C^{\ast }$ algebra.

(i)	$\Vert a\Vert =\Vert a^{\ast }\Vert$.
(ii)	For $a\in \mathrm{Re}\mathcal{A}$, $\Vert a\Vert =r(a)$.
(iii)	$\rho :\mathcal{A}\to \mathcal{B}$ is a $\ast$-homo, then $\Vert \rho (a)\Vert \le \Vert a\Vert$. (Coro: $\rho$ is $\ast$-iso implies isometry.)
(iv)	For unital abelian $\mathcal{A}$ and $h:\mathcal{A}\to \mathbb{C}$, $\Vert h\Vert =1$.
(v)	For $a\in \mathrm{Re}\mathcal{A}$, $h(a)\in \mathbb{R}$. (Coro: $h(a^{\ast })=\overline{h(a)}$.)
(vi)	If $\mathcal{A}$ is unital abelian, $\Lambda :\mathcal{A}\to C(\Delta )$ is a $\ast$-isomorphism. $\square$

证明.

(i)	$\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, and $a^{\ast \ast }=a$.
(ii)	$\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert =\Vert a^2\Vert$, so $r(a)=\underset{n\to \infty }{\lim }\Vert a^{2^n}{\Vert }^{1/2^n}=\Vert a\Vert$.
(iii)	By extending $\rho$ when necessary, we can assume $\mathcal{A}$ is unital and $\rho (1)=1$. Note that $\sigma (\rho (a))\subseteq \sigma (a)$, then $\Vert \rho (a){\Vert }^2=\Vert \rho (a^{\ast }a)\Vert =r(\rho (a^{\ast }a))\le r(a^{\ast }a)=\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.
(iv)	If there is $\Vert a\Vert /h(a)<1$, then exists $b=(1-a/h(a){)}^{-1}$, and it follows that $1=h(b(1-a/h(a))=0$.
(v)	Consider the abelian $C^{\ast }$ algebra generated by $a+it$ and $1$, where $t\in \mathbb{R}$. Then $$|h(a+it){|}^2\le \Vert a+it{\Vert }^2=\Vert a^2+t^2\Vert \le \Vert a^2\Vert +t^2.$$Write $h(a)=\alpha +i\beta$, then $|h(a+it){|}^2=|\alpha {|}^2+(\beta +t{)}^2$, forcing $\beta =0$.
(vi)	When $a\in \mathrm{Re}\mathcal{A}$, $\Vert \hat{a}{\Vert }_{\infty }=r(a)=\Vert a\Vert$. Hence for any $a$, $\Vert \hat{a}{\Vert }_{\infty }^2=\Vert {\hat{a}}^{\ast }\hat{a}{\Vert }_{\infty }=\Vert \widehat{a^{\ast }a}{\Vert }_{\infty }=\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2.$ Moreover, as $\Lambda (\mathcal{A})$ is a closed algebra separating points of $\Delta$, $\Lambda$ is surjective by Stone–Weierstrass. $\square$

Theo.

(i)	For $a\in \mathrm{Re}\mathcal{A}$, $\sigma (a)\subseteq \mathbb{R}$.
(ii)	For two $C^{\ast }$ algebras $\mathcal{B}\subseteq \mathcal{A}$ with common identity, ${\sigma }_{\mathcal{B}}(a)={\sigma }_{\mathcal{A}}(a)$.

证明.

(i)

If $\mathcal{A}$ is abelian, it follows from (v) of the preceding proposition. If $\mathcal{A}$ is not, nonetheless $C^{\ast }(a)$ is abelian, then ${\sigma }_{\mathcal{A}}(a)\subseteq {\sigma }_{C^{\ast }(a)}(a)\subseteq \mathbb{R}$.

(ii)

When $a\in \mathrm{Re}\mathcal{B}$, ${\sigma }_{C^{\ast }(a)}\subseteq \mathbb{R}$ and $\partial {\sigma }_{C^{\ast }(a)}(a)\subseteq \partial {\sigma }_{\mathcal{A}}(a)$ implies that ${\sigma }_{\mathcal{B}}(a)={\sigma }_{\mathcal{A}}(a)$. Now for $a\in \mathcal{B}$, suppose that $a$ is invertible in $\mathcal{A}$, i.e. $ax=xa=1$, then $(a^{\ast }a)(x{x}^{\ast })=(x{x}^{\ast })(a^{\ast }a)=1$, where $a^{\ast }a\in \mathrm{Re}\mathcal{B}$ and is thus invertible in $\mathcal{B}$ (as we have shown in the previous step that ${\sigma }_{\mathcal{B}}(a^{\ast }a)={\sigma }_{\mathcal{A}}(a^{\ast }a)$), so $x{x}^{\ast }\in \mathcal{B}$ by the uniqueness of inverse, then $x=(x{x}^{\ast })a^{\ast }\in \mathcal{B}$. $\square$

Theo. Let $\Vert f{\Vert }_{E,\infty }=\underset{E(S)=0}{\inf }\underset{x\in \Omega \setminus S}{\sup }|f(x)|$, then $$\pi :L^{\infty }(\Omega ,E)\to \mathcal{B}(H),f\mapsto \int f\mathrm{d}E$$is an isometric $C^{\ast }$-homo.

Theo. If $\mathcal{A}$ is a $C^{\ast }$ algebra, $a$ normal and $f\in C(\sigma (a))$, then $\sigma (f(a))=f(\sigma (a))$.

证明. The map $f\mapsto f(a)$ is a $\ast$-isomorphism from $C(\sigma (a))$ to $C^{\ast }(a)$, so $\sigma (f(a))={\sigma }_{C(\sigma (a))}(f)=f(\sigma (a))$.

$\square$

Prop. $\phi \in {\mathcal{A}}^{\ast }$ is positive iff $\Vert \phi \Vert =\phi (1)$.

证明. For positive $\phi$, $\Vert x\Vert \le 1$ implies $x^{\ast }x\le 1$ and hence $$|\phi (x)|\le \phi (1{)}^{1/2}\phi (x^{\ast }x{)}^{1/2}\le \phi (1).$$

For the converse, if $\mathcal{A}$ is abelian, $\phi$ can be seen as in the form $C(X)\to \mathbb{C},f\mapsto {\int }_{X}f\mathrm{d}\mu$. Then if $\Vert \mu \Vert =\mu (X)$, $\mu$ would have to be positive.

And if $\mathcal{A}$ is not abelian, for $a\in {\mathcal{A}}_{+}$ consider the abelian $C^{\ast }(a)$, then $$\phi {|}_{C^{\ast }(a)}(1)\le \Vert \phi {|}_{C^{\ast }(a)}\Vert \le \Vert \phi \Vert =\phi (1)=\phi {|}_{C^{\ast }(a)}(1),$$so $\phi {|}_{C^{\ast }(a)}$ is positive.

$\square$

Prop. $T\subset \mathcal{B}(X)$ with $\sigma (T)=F_1\cup F_2$, where $F_1,F_2$ are disjoint nonempty closed sets. Let $G_1,G_2$ be disjoint open sets containing $F_1,F_2$. Let $E_i={\chi }_{G_i}(T)$ and $X_i=E_{i}X$.

(i) $X=X_1+X_2$, and $ST=TS$ implies $S{X}_i\subset X_i$;

(ii) let $T_i=T{|}_{X_i}$, then $\sigma (T_i)=F_i$;

(iii) let $R:X_1+X_2\to X_1\oplus X_2$, then $R$ is invertible and $RT{R}^{-1}=T_1\oplus T_2$.

证明. (i) ${\chi }_{G_i}\in \mathrm{Hol}(T)$, so $S{E}_i=E_{i}S$, then $S{X}_i=S{E}_{i}X=E_{i}SX\subset E_{i}X=X_i$;

(ii) Let $g(z)=(z-\lambda {)}^{-1}$ for some $\lambda \notin G_1$ on $G_1$ and $0$ on $G_2$, then $g\in \mathrm{Hol}(T)$ with $$g(T)=\frac{1}{2\pi \mathrm{i}}{\int }_{\Gamma }\frac{g(z)}{z-T}\mathrm{d}z,$$where $\Gamma \subset G_1$, and $F_1\subseteq \mathrm{Ins}\Gamma$. Then $$E_1(T-\lambda )E_{1}g(T)E_1=E_{1}g(T)E_1(T-\lambda )E_1=E_1,$$put $A=E_{1}g(T){|}_{X_1}$, then $(T_1-\lambda )A=A(T_1-\lambda )=I_{X_1}$, so $\lambda \in \rho (T_1)$. Hence $\sigma (T_1)\subseteq F_1$. Moreover, $\sigma (T_1)\cup \sigma (T_2)=\sigma (T)=F_1\cup F_2$, so $\sigma (T_i)=F_i$.

$\square$

Theo. $\phi$ is a positive linear functional on $\mathcal{A}$, then there is a cyclic representation $(\pi ,H)$ st $\phi (a)=⟨\pi (a)e,e⟩$.

证明. When $\mathcal{A}=C(X)$ for compact $X$, then $\phi (f)={\int }_{X}f\mathrm{d}\mu$. Let $H=L^2(\mu )=$ the completion of $C(X)/\mathcal{L}$ where $$\mathcal{L}=\{f\in C(X):{\int }_X|f{|}^2\mathrm{d}\mu =0\},⟨[f],[g]⟩={\int }_{X}f\bar{g}\mathrm{d}\mu .$$More generally, let $H$ be the completion of $\mathcal{A}/\mathcal{L}$ where $$\mathcal{L}=\{x\in \mathcal{A}:\phi (x^{\ast }x)=0\},⟨[x],[y]⟩=\phi (y^{\ast }x).$$Let $\pi$ be the operator on $\mathcal{A}/\mathcal{L}$ such that $\pi (a)[x]=[ax]$, then as $\Vert \pi (a)\Vert \le \Vert a\Vert$, $\pi (a)$ can be extended as an operator on $H$. Let $e=[1]$, then $\pi (\mathcal{A})e=\mathcal{A}/\mathcal{L}$ is by definition dense in $H$, and $$⟨\pi (a)e,e⟩=⟨[a],[1]⟩=\phi (a).$$Suppose that $({\pi }^{\prime },H^{\prime })$ is another representation. Let $$U:\pi (\mathcal{A})e\to {\pi }^{\prime }(\mathcal{A})e^{\prime },\pi (a)e\mapsto {\pi }^{\prime }(a)e^{\prime }.$$Extend the isometry $U$ to $H\to H^{\prime }$, then $$U\pi (a)\pi (x)e=U\pi (ax)e={\pi }^{\prime }(ax)e^{\prime }={\pi }^{\prime }(a){\pi }^{\prime }(x)e^{\prime }={\pi }^{\prime }(a)U\pi (x)e,$$so $U\pi (a)={\pi }^{\prime }(a)U$.

$\square$

证明. Let $K(a,b)=\phi (b^{\ast }a)$, then $K$ is positive definite, i.e. $$\sum _{i,j}{\alpha }_i{\bar{\alpha }}_{j}K(b_i,b_j)\ge 0.$$Write $K_a(b)=K(a,b)$, and $\mathcal{F}=\mathrm{span}\{K_a\}$, and define the inner product (where $f={\sum }_i{\alpha }_i{K}_{b_i}$) $$[f,f]=\sum _{i,j}{\alpha }_i{\bar{\alpha }}_{j}K(b_i,b_j).$$Note that $$\phi ((af{)}^{\ast }(af))=\phi (f^{\ast }{a}^{\ast }af)\le \phi (f^{\ast }\Vert a{\Vert }^{2}f)=\Vert a{\Vert }^2\phi (f^{\ast }f).$$Therefore, if we let $\pi (a)$ be the operator on $\mathcal{F}$ st $$\pi (a)\left(\sum _i{\alpha }_i{K}_{b_i}\right)=\sum _i{\alpha }_i{K}_{a{b}_i},$$then as $$\left[\sum _i{\alpha }_i{K}_{b_i},\sum _i{\alpha }_i{K}_{b_i}\right]=\sum _{i,j}{\alpha }_i{\bar{\alpha }}_j\phi (b_i,b_j)=\phi (f^{\ast }f),$$and consequently $$\left[\sum _i{\alpha }_i{K}_{a{b}_i},\sum _i{\alpha }_i{K}_{a{b}_i}\right]=\phi ((af{)}^{\ast }af)\le \Vert a{\Vert }^2\phi (f^{\ast }f)=\Vert a{\Vert }^2\left[\sum _i{\alpha }_i{K}_{b_i},\sum _i{\alpha }_i{K}_{b_i}\right],$$$\pi$ is a bounded operator and can thus be extended as an operator on $H$, the completion of $\mathcal{F}$ under its inner product. It is trivial to see that $\pi (ab)=\pi (a)\pi (b)$, and $$\left[\sum _i{\alpha }_i{K}_{a{b}_i},\sum _j{\beta }_i{K}_{c_j}\right]=\sum _{i,j}{\alpha }_i{\bar{\beta }}_j\phi (c_j^{\ast }a{b}_i)=\sum _{i,j}{\alpha }_i{\bar{\beta }}_j\phi ((a^{\ast }{c}_j{)}^{\ast }{b}_i)=\left[\sum _i{\alpha }_i{K}_{b_i},\sum _j{\beta }_j{K}_{a^{\ast }{c}_j}\right]$$gives ${\pi }^{\ast }(a)=\pi (a^{\ast })$. Finally, $[\pi (a)K_e,K_e]=K(a,e)=\phi (a)$.

$\square$

Theo. If $X$ is compact Hausdorff and $\rho :C(X)\to \mathcal{B}(H)$ is a $C^{\ast }$-homo that maps $1$ to $I$, then there exists a unique spectral measure $(E,H)$ st $$\pi (f)=\int f\mathrm{d}E$$and (i) $\pi$ is injective iff for all nonempty open $V$, $E(V)\ne 0$, (ii) $S\pi (f)=\pi (f)S$ iff $SE(\omega )=E(\omega )S$.

证明. For given $x,y\in H$, by Riesz we have a complex regular Borel measure ${\mu }_{x,y}$ st $$⟨\pi (f)x,y⟩=\int f\mathrm{d}{\mu }_{x,y}$$and (i) ${\mu }_{x,y}={\bar{\mu }}_{y,x}$, (ii) $\Vert {\mu }_{x,y}\Vert \le \Vert x\Vert \Vert y\Vert$. Then there is an operator $\tilde{\pi }(\phi )$ for every bounded Borel function $\phi$ st $$⟨\tilde{\pi }(\phi )x,y⟩=\int \phi \mathrm{d}{\mu }_{x,y}.$$From $$\int fg\mathrm{d}{\mu }_{x,y}=⟨\pi (fg)x,y⟩=⟨\pi (f)\pi (g)x,y⟩=\int f\mathrm{d}{\mu }_{\pi (g)x,y}$$we get $g\mathrm{d}{\mu }_{x,y}=\mathrm{d}{\mu }_{\pi (g)x,y}$. And from $$\int \phi g\mathrm{d}{\mu }_{x,y}=⟨\tilde{\pi }(\phi )\pi (g)x,y⟩=⟨\pi (g)x,\tilde{\pi }(\phi )y⟩=\int g\mathrm{d}{\mu }_{x,{\tilde{\pi }}^{\ast }(\phi )y}$$we get $\phi \mathrm{d}{\mu }_{x,y}=\mathrm{d}{\mu }_{x,{\tilde{\pi }}^{\ast }(\phi )y}$. Then $$\int \phi \psi \mathrm{d}{\mu }_{x,y}=\int \psi \mathrm{d}{\mu }_{x,{\tilde{\pi }}^{\ast }(\phi )y}=⟨\tilde{\pi }(\psi )x,{\tilde{\pi }}^{\ast }(\phi )y⟩=⟨\tilde{\pi }(\phi \psi )x,y⟩,$$so $\tilde{\pi }(\phi \psi )=\tilde{\pi }(\phi )\tilde{\pi }(\psi )$. And (i) implies ${\tilde{\pi }}^{\ast }(\phi )=\tilde{\pi }(\bar{\phi })$. Therefore $\tilde{\pi }$ is a $C^{\ast }$-extension of $\pi$ from $C(X)$ to bounded Borel functions.

Now define $E(\omega )=\tilde{\pi }({\chi }_{\omega })$. Easy to check that $E(\varnothing )=0,E(X)=I,E({\omega }_1\cap {\omega }_2)=E({\omega }_1)E({\omega }_2)$, and $${\Vert E(\omega )x-\sum _{i=1}^{n}E({\omega }_i)x\Vert }^2=⟨E(\omega \setminus {\cup }_{i=1}^n{\omega }_i)x,x⟩={\mu }_{x,x}(\omega \setminus {\cup }_{i=1}^n{\omega }_i)x,x)\to 0.$$As $$⟨E(\omega )x,y⟩=⟨\tilde{\pi }({\chi }_{\omega })x,y⟩=\int {\chi }_{\omega }\mathrm{d}{\mu }_{x,y}={\mu }_{x,y}(\omega ),$$we have $$⟨\pi (f)x,y⟩=\int f\mathrm{d}{\mu }_{x,y}=\int f\mathrm{d}{E}_{x,y}.$$(i) If $E(V)=0$ for some nonempty open $V$, then for continuous function $f\ne 0$ supported on $V$, $\pi (f)=\int f\mathrm{d}E=0$, so $\pi$ is not injective.

If $\pi$ is not injective, then the closed ideal $\ker \pi$ is of the form $\{f:f{|}_S=0\}$. Pick an open $U\cap S=\varnothing$, then a continuous function $g\ge 0$ supported on $U$ belongs to $\ker \pi$, so $\int g\mathrm{d}E=\pi (g)=0$, implying $E(U)=0$.

(ii) Note that $$⟨S\pi (f)x,y⟩=⟨\pi (f)Sx,y⟩\iff \int f\mathrm{d}{E}_{x,S^{\ast }y}=\int f\mathrm{d}{E}_{Sx,y}$$and $$⟨SE(\omega )x,y⟩=⟨E(\omega )Sx,y⟩\iff E_{x,S^{\ast }y}(\omega )=E_{Sx,y}(\omega ),$$so $S\pi (f)=\pi (f)S$ iff $SE(\omega )=E(\omega )S$ for each $\omega$.

$\square$

Theo. $\mathbf{N}=\{N_1,\cdots ,N_m\}$ is a set of commuting normal operators, generating with $I$ a closed self–adjoint subalgebra $C^{\ast }(\mathbf{N})$. Let $M$ be the maximal ideal space of $C^{\ast }(\mathbf{N})$, then $$\tau :M\to \Delta \subset {\mathbb{C}}^m,\phi \mapsto (\phi (N_1),\cdots ,\phi (N_m))$$is a homeomorphism, so $$\pi :C(\Delta )\to C^{\ast }(\mathbf{N}),f(z_1,\cdots ,z_m)\mapsto (f(N_1),\cdots ,f(N_m))$$is a $C^{\ast }$-isomorphism. There is unique spectral measure $(E,H)$ on $\Delta$ st $$N_k={\int }_{\Delta }{z}_k\mathrm{d}E,k=1,\cdots ,m,$$and (i) $T$ commutes with each $N_k$ iff $TE(\omega )=E(\omega )T$, (ii) each $E(\omega )$ is a reduced subspace for each $N_k$, (iii) $N_{k}h={\lambda }_{k}h$ for some $h\ne 0$ iff $E(\{({\lambda }_1,\cdots ,{\lambda }_m)\})\ne 0$.

证明. We have established that there exists $(E,H)$ on $\Delta$ st $N_k={\int }_{\Delta }{z}_k\mathrm{d}E$. Then $$p(N_1,N_1^{\ast },\cdots ,N_m,N_m^{\ast })={\int }_{\Delta }p(z_1,{\bar{z}}_1,\cdots ,z_m,{\bar{z}}_m)\mathrm{d}E,$$for any polynomial $p$, and hence for continuous functions.

(i) is by (ii) above, (ii) is due to $E(\omega )N_k=N_{k}E(\omega )$, and (iii) is by (i) above.

$\square$

Theo. Let $N$ be a normal operator, then there is a spectral measure $E$ defined on the Borel subsets of $\sigma (N)$ st for $f\in C(\sigma (N))$, $$f(N)={\int }_{\sigma (N)}z\mathrm{d}E(z)$$with: (i) $\lambda$ is an eigenvalue of $N$ iff $E(\{\lambda \})\ne 0$; (ii) $TN=NT$ iff $TE(\omega )=E(\omega )T$ for all Borel $\omega$.

证明. (i) For eigenvalue $\lambda$, write $Nx=\lambda x$. Let $$f_n(z)=\{\begin{array}{ll}1/(\lambda -z),& z\notin B_{1/n}(\lambda ),\\ 0,& z\in B_{1/n}(\lambda ).\end{array}$$Then $$E(\mathbb{C}\setminus B_{1/n}(\lambda ))x=f_n(N)(\lambda -N)x=0,$$letting $n\to \infty$ gives $E(\mathbb{C}\setminus \{\lambda \})x=0$, i.e. $E(\{\lambda \})x=x$.

(ii) By Fuglede, $T{N}^{\ast }=N^{\ast }T$, so $Tp(N,N^{\ast })=p(N,N^{\ast })T$ for any polynomial, and subsequently $Tf(N)=f(N)T$ for any $f\in C(\sigma (N))$, and therefore $TE(\omega )=E(\omega )T$ for any open set $\omega$ by taking a sequence of nonnegative continuous $f_n\nearrow 1_{\omega }$, and ultimately for any Borel $\omega$. For the converse, approximate $z$ by simple functions.

$\square$

Theo. If $\{U_t\}$ is a weakly continuous one parameter unitary group, then there is unique spectral measure $(P,H)$ st $$U_t={\int }_{\mathbb{R}}{\mathrm{e}}^{-\mathrm{i}st}\mathrm{d}P(s).$$

Theo. Let $\mathcal{A}$ be a self–adjoint algebra of $B(H)$ with $\mathcal{A}v=0\Rightarrow v=0$. Let ${\mathcal{A}}_s$ and ${\mathcal{A}}_w$ be the strong and weak closures of $\mathcal{A}$. Then ${\mathcal{A}}_s={\mathcal{A}}_w={\mathcal{A}}^{\prime \prime }$.

证明. Clearly ${\mathcal{A}}_s\subseteq {\mathcal{A}}_w\subseteq {\mathcal{A}}^{\prime \prime }$, since if there is a sequence of $\{A_i\}$ in $\mathcal{A}$ st $⟨A_{i}x,y⟩\to ⟨Ax,y⟩$, then it follows from $⟨ASx,y⟩\leftarrow ⟨A_{i}Sx,y⟩=⟨S{A}_{i}x,y⟩\to ⟨SAx,y⟩$ that $AS=SA$ for all $S\in {\mathcal{A}}^{\prime }$. Then it suffices to show that every operator $B\in {\mathcal{A}}^{\prime \prime }$ can be strongly approximated by operators in $\mathcal{A}$; that is, to find $A\in \mathcal{A}$ st $\sum _{k=1}^n\Vert (B-A)h_k\Vert <\epsilon$.

Consider first the case $n=1$, and let $P$ be the projection onto $\overline{\mathcal{A}h}$. Since for all $A\in \mathcal{A}$ we have $A\overline{\mathcal{A}h}\subset \overline{\mathcal{A}h}$ and $A({\overline{\mathcal{A}h}}^{\perp })\subset {\overline{\mathcal{A}h}}^{\perp }$, $AP=PA$ so $P\in {\mathcal{A}}^{\prime }$. Next observe that $h\in \overline{\mathcal{A}h}$, for $A(I-P)h=(I-P)Ah=0$ implies $(I-P)h=0$. Now $BP=PB$ implies $Bh=BPh=PBh\in \overline{\mathcal{A}h}$.

Then consider the case $n\ge 2$. Let $H_n=H\oplus \cdots \oplus H$ be the direct sum of $n$ copies of $H$, and let $h=(h_1,\cdots ,h_n)$. And let ${\mathcal{A}}_n=\mathcal{A}\oplus \cdots \oplus \mathcal{A}$ and $B_n=B\oplus \cdots \oplus B$, then simple calculation indicates that $({\mathcal{A}}_n{)}^{\prime }=\{(T_{ij}):T_{ij}\in {\mathcal{A}}^{\prime }\}$ and that $({\mathcal{A}}_n{)}^{\prime \prime }=({\mathcal{A}}^{\prime \prime }{)}_n$. So $B_{n}h\in (A_n{)}^{\prime \prime }h=\overline{{\mathcal{A}}_{n}h}$.

$\square$